A certain circle can be represented by the following equation. $x^2+y^2+8x-16y+31=0$ What is the center of this circle ? $($
Answer: The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2+y^2+8x-16y+31&=0\\\\ x^2+y^2+8x-16y&=-31\\\\ (x^2+8x)+(y^2-16y)&=-31 \text{(rearrange terms)}\\\\ (x^2+8x{+16})+(y^2-16y{+64})&=-31{+16}{+64}\end{aligned}$ Notice that we must add ${16}$ and ${64}$ on the right side of the equation, since we added them to the left side of the equation. [How did we get 16 and 64?] Writing the equation in standard form $\begin{aligned}(x^2+8x{+16})+(y^2-16y{+64})&=-31{+16}{+64}\\\\ (x+4)^2+(y-8)^2&=49\\\\ (x-(-4))^2+(y-8)^2&=7^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(-4,8)$ and has a radius of $7$ units. Summary The circle is centered at $(-4,8)$. The circle has a radius of $7$ units.